[proofplan] We proceed by strong induction on the length $\ell$ of the odd circuit. If the vertices of the circuit are all distinct, the circuit is itself an odd cycle and we are done. Otherwise the circuit revisits some vertex, and we cut it at the repeated vertex into two shorter subcircuits whose lengths sum to $\ell$. Since $\ell$ is odd, exactly one of the subcircuits is odd, and strong induction on that shorter odd subcircuit produces the desired odd cycle. [/proofplan] [step:Set up strong induction on the length of the odd circuit] Let $C$ be an odd circuit in $G$, written as a sequence of vertices \begin{align*} C: x_1, x_2, \ldots, x_\ell, x_1 \end{align*} where each consecutive pair $\{x_i, x_{i+1}\}$ is an edge of $G$ (indices read modulo $\ell$) and $\ell \ge 1$ is odd. Here $\ell$ is the length of $C$, i.e., the number of edges traversed. We prove the statement "every odd circuit of length $\ell$ in $G$ contains an odd cycle" by strong induction on $\ell$. [guided] Recall that a **circuit** is a closed walk: a sequence of vertices $x_1, x_2, \ldots, x_\ell, x_1$ such that each consecutive pair is an edge, with no requirement that the vertices be distinct. A **cycle**, by contrast, is a circuit in which the only repeated vertex is the first/last one. The theorem claims that although circuits may revisit vertices, any odd circuit must contain an honest cycle of odd length as a subgraph. Why induct on $\ell$? The argument below will reduce an odd circuit that repeats a vertex to a strictly shorter odd circuit inside the same graph. Strong induction is the correct form because the shorter circuit could have any length strictly less than $\ell$, not just $\ell - 1$. The base cases are small odd lengths. The minimum length of a circuit in a simple graph is $\ell = 3$ (since loops $\ell = 1$ and repeated edges $\ell = 2$ are excluded). A circuit of length $3$ has the form $x_1, x_2, x_3, x_1$; if this is a valid circuit in a simple graph, the vertices $x_1, x_2, x_3$ must be pairwise distinct, so the circuit is already a $3$-cycle (a triangle). This settles $\ell = 3$. Larger odd lengths are handled by the inductive step below. [/guided] [/step] [step:Handle the case where all vertices of the circuit are distinct] Suppose first that $x_1, \ldots, x_\ell$ are pairwise distinct. Then by definition $C$ is an odd cycle in $G$, and $C$ itself is the desired odd cycle contained in $C$. [/step] [step:Split the circuit at a repeated vertex into two shorter subcircuits] Suppose instead that the vertices of $C$ are not all distinct. Then there exist indices $1 \le i < j \le \ell$ with $x_i = x_j$. Define two subcircuits: \begin{align*} C_1 &: x_i, x_{i+1}, \ldots, x_{j-1}, x_j, \qquad \text{of length } j - i, \\ C_2 &: x_j, x_{j+1}, \ldots, x_\ell, x_1, \ldots, x_{i-1}, x_i, \qquad \text{of length } \ell - (j - i). \end{align*} Both $C_1$ and $C_2$ are closed walks in $G$ because their endpoints agree ($x_i = x_j$), and each uses only edges from $C$, hence edges of $G$. Thus $C_1$ and $C_2$ are circuits in $G$. Let $\ell_1 := j - i$ and $\ell_2 := \ell - (j - i)$, so $\ell_1 + \ell_2 = \ell$. Since $1 \le i < j \le \ell$, we have $1 \le \ell_1 \le \ell - 1$ and $1 \le \ell_2 \le \ell - 1$; in particular both subcircuits are strictly shorter than $C$. [guided] The geometric picture is clear: the circuit $C$ traces a closed walk that returns to some vertex $x_i = x_j$ before completing its full loop. Cutting at that shared vertex gives two closed sub-walks, $C_1$ (the "inner loop" from $x_i$ back to itself via $x_j$) and $C_2$ (the "outer loop" going the other way round). Each is a valid circuit in $G$ because cutting and re-gluing at a common vertex preserves the walk property. The key fact is that both pieces are strictly shorter than $C$: $C_1$ has length $\ell_1 = j - i \ge 1$, and $C_2$ has length $\ell_2 = \ell - (j - i) \ge 1$, since $1 \le i < j \le \ell$ forces $1 \le j - i \le \ell - 1$. Note also that at this stage we make no claim about whether $C_1$ or $C_2$ is a valid circuit in the sense of being non-trivial: even a trivial "circuit" of length $1$ (a loop) or length $2$ (a repeated edge) would, in a simple graph, force $\ell_1 \ge 3$ or $\ell_2 \ge 3$. The parity argument in the next step does not depend on this. [/guided] [/step] [step:Identify the odd subcircuit via parity of $\ell_1 + \ell_2 = \ell$] Since $\ell_1 + \ell_2 = \ell$ and $\ell$ is odd, exactly one of $\ell_1, \ell_2$ is odd and the other is even. Let $C^* \in \{C_1, C_2\}$ be the subcircuit with odd length $\ell^* \in \{\ell_1, \ell_2\}$. Then $C^*$ is an odd circuit in $G$ with $\ell^* < \ell$. [/step] [step:Apply the inductive hypothesis to the shorter odd subcircuit] By the strong inductive hypothesis applied to $C^*$ — an odd circuit in $G$ of length $\ell^* < \ell$ — there exists an odd cycle contained in $C^*$. Since the edges of $C^*$ are a subset of the edges of $C$, this odd cycle is also contained in $C$. This completes the inductive step and hence the proof. [/step]